Question

A baseball is hit at an initial speed of 40 m/s at an angle of 60° above the horizontal. The ball flies for 10 s.

Find the distance, in meters, that the ball travels. Use g = 9.81 m/s2.
A) 100 m
B) 146 m
C) 200 m
D) 346 m

Answer 1

initial velocity=u=40 m/s

a=g=9.8m/s2

angle of projection=θ=60°

A) maximum Height h= u²sin²θ/2g

h= 40x40 sin60²/2x10

h= 40x40 x3/4x2 x 10

h=60m

B) Horizontal Range = X=u²sin2θ/g

=X= 40x40 sin2x60/10

=40x40 x√3/2x10

=80√3 m

Answer 2

c) x = 200 m

Step-by-step explanation:

Ball is projected at speed of 40 m/s at an angle of 60 degree

so here we have two components of velocity given as

`v_x = 40 cos60`

`v_x = 20 m/s`

`v_y = 40 sin60`

`v_y = 34.64 m/s`

now the horizontal distance moved by the ball in next 10 s is given as

`x = v_x * t`

so we have

`x = 20 (10)`

`x = 200 m`